Question about commutators

[Lotsofsloths]

How would you simulate a J perm using a commutator?
J Perm = R U R' F' R U R' U' R' F R2 U' R' U'

 

[Johannes91]

There's a typo in your alg. And it's not possible, because J-perm has odd parity and a commutator has an even number of moves in QTM. You can do a 3-cycle, though. Here's one way:

[x' r2 : [F U2 F', d2]] = L2 D F2 D' F2 U F2 U' F2 L2

 

[AvGalen]

I have no idea for an actual alg, but this is how I would search for 1:

Most commutators are a 3-cycle of something, mostly corners, sometimes edges, but there is no reason why you couldn't make it a 3-cycle of 3 corner-edge-pairs.
If you look closely at the J-perm you see that it switches 2 corners and 2 edges. But with a U-turn you can make that 3 corners and 3 edges.
Now "all you have to do" is find some way to position the 3 corner-edge-pairs, figure out the actual commutator to cycle the pairs and finally undo the first moves.

I think I have actually done this for an FMC-solve at least 6 months ago (around the time I got my first 25 moves solution). The pairs were already in the correct position during the beginning of the OLL so I didn't have to use setupmoves. The commutator was 8 moves, but this is all I remember.

Why would you want this?

 

[MistArts]

Why find a commutator? Just use the optimal alg.

 

[Johannes91]

Why find a commutator?

Because he wants to understand what he's doing?

 

[ch_ts]

Well, the 2nd J-perm listed here:
http://www.ws.binghamton.edu/fridrich/Mike/permute.html
(U'FB U2B'U'B U2F'U B')

can be done explained like so:

U'BFd2F'U'Fd2F'UB'
or
U' - to make it so there are 3 corner-edge pairs to cycle
B - set up
[Fd2F',U'] - do the cycle
B' - undo setup

 

[Swordsman Kirby]

The J perm was probably the first commutator I found that moved both corners and edges back in the day:

[L U' R d2 R' U L',U'] U

 

[Lotsofsloths]

I have no idea for an actual alg, but this is how I would search for 1:

Most commutators are a 3-cycle of something, mostly corners, sometimes edges, but there is no reason why you couldn't make it a 3-cycle of 3 corner-edge-pairs.
If you look closely at the J-perm you see that it switches 2 corners and 2 edges. But with a U-turn you can make that 3 corners and 3 edges.
Now "all you have to do" is find some way to position the 3 corner-edge-pairs, figure out the actual commutator to cycle the pairs and finally undo the first moves.

I think I have actually done this for an FMC-solve at least 6 months ago (around the time I got my first 25 moves solution). The pairs were already in the correct position during the beginning of the OLL so I didn't have to use setupmoves. The commutator was 8 moves, but this is all I remember.

Why would you want this?

I was just wondering how you would solve a case like that with a commutator.. but I guess all I needs was U'r advice xDDDDDD

 

[ShadenSmith]

Wow, that was a really lame joke. I love lame jokes.

 

[AvGalen]

I found that FMC solution. It was in 2008-5:

3x3x3 Fewest Moves
1. B F2 D F2 L' R' B F' R2 F' U' B' D' U L R2 B F' D2 L' D2 U R B' L2 R D' R2 D' U2 B2 L2 R' U' F2 L' R F L R2 F L2 D' U2 L2 (45 moves original)
R2 D2 R2 B' D L' R U' L B2 R2 B' D F R' B D F (18 moves optimal solution, easier to apply multiple times)

Improvement 2, final solution (27): F2 D2 B D' L U2 D B D B2 L' B D' B D2 B2 R' B D' B' R B D R' D R D
x-cross + prepare pair: F2 D2 B D' L U2
double x-cross + prepare last pairs: D B D B2 L' B D'
insert last pairs: B D2 B' . R' D R D
fix last layer at . : B' R' B D' B' R B D

And this is the commutator I was looking for: (B' R' B) D' (B' R B) D